Direct limits of injective modulesSee page 81 Theorem 3.46 Lectures on Modules and Rings by Lam. This has exactly the theorem refered to in the comment below and is a nice textbook.

Nov1

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Pure Subgroups of $\mathbb Z\times\mathbb Z$The subgroup $\langle (1,2),(3,2)\rangle$ contains $(2,0)$. It does not contain $(1,0)$ since if it did there would exist $n,m\in\mathbb{N}$ such that $1=n+3m$ and $0=2n+2m$. But this would mean $1=2m$. You don't need any tools to do this question.

Transcendental extensions of $\mathbb{Q}$ containing algebraic elements.I think maybe I get it now. If we take $f\in\mathbb{Q}[x]$ to be the minimal polynomial of the primitive element of $\mathbb{Q}(a,b)$ then $f$ does not become reducible over $F$ but does become reducible over $K(v)$. Thus, $[K(v):F]$ is the degree of $f$ and so is $[K(v):\mathbb{Q}(v)]$. Thus $[F:\mathbb{Q}(v)]=1$.

Oct19

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Transcendental extensions of $\mathbb{Q}$ containing algebraic elements.I don't understand the line "Then by (1) $\[F:\mathbb{Q}(v)\]=1$" i.e. $F$ might contain no elements algebraic over $\mathbb{Q}$ but still contain elements algebraic over $\mathbb{Q}(v)$. Could you explain this? Thanks for the answer. I think I can construct my own proof (which is heavily inspired by your proof) but I'd like to understand this step in your proof (and then accept your answer!).